1

Integrate by parts to prove the interpolation inequality: $$ ||Du||_{L_2}\leq C||u||^{1/2}_{L^2} ||D^2 u||^{1/2}_{L^2}, $$ for all $u \in C_0^{\infty}(U),$ where $U$ is some open subset of $\mathbb{R}^n.$

I am incredibly weak in multivariable situations and have no idea what to do any help would be appreciated.

It would be extremely helpful if somebody knew of a place where I could read what the integration of parts formula was in this situation.

User112358
  • 1,537

1 Answers1

2

Presumably what it's getting at is this: $$ \lVert Du \rVert_2^2 = \int_U \partial_i u \partial_i u = -\int_U u \partial_i \partial_i u, $$ using the fact that $u \in C_0^{\infty}(U)$ to discard the boundary terms (there are $n$ of them, and they all look like $[ u \partial_i u ]_{\partial U}$.) Then applying Cauchy-Schwarz, $$ \left\lvert \int_U (-u) \partial_i \partial_i u \right\rvert \leqslant \lVert -u \rVert_2 \left\lvert \int_U (\partial_i \partial_i u)^2 \right\rvert^{1/2} $$ The last factor's integrand is smaller than $$ \lvert D^2 u \rvert^2 = \sum_{i,j} (\partial_i \partial_j u)^2, $$ and hence $$ \lVert Du \rVert_2^2 \leqslant \lVert u \rVert_2 \lVert D^2 u \rVert_2, $$ and the result follows by taking a square root.

Chappers
  • 67,606
  • Thanks! I was really lost as to what was meant by the integration by parts formula since our integral involved $Du$. So this helped. – User112358 Sep 14 '15 at 01:08
  • In the first line the LHS is always positive since it's a norm, but the RHS can be negative if $u$ is positive everywhere. – Math_Day Jun 07 '22 at 22:03